Result 1 If `V_0` is the value of an article at a certain time and `R%` per annum is the rate of depreciation then the value `V_n` at the end of n years is given by `V_n=V_0 (1- R/1000)^n`

Ressult 1 If V_(0) is the value of an article at a certain time and R% per annum is the rate of depreciation then the value V_(n) at the end of n years is given by V_(n)=V_(0)(1-(R)/(1000^(n))

Result 2 If V_0 is the value of an article at a certain time and the rate of depreciation is R_1 % for first n_1 years r_2 % for next n_2 years and so on and R_k % for the last n_k years then the V=V_0(1- R_1/100)^(n1) (1- R_2/100)^(n2) .......... (1- R_3/100)^(nk)

The value of a machien being decreased in y years at the rate of r% per year, becomes Rs. V. Find the value of the machine before n years.

Formula 1 Let P be the principal and the rate of interest be R% per annum.If the interest is compunded anually then the amount A and the compound interest C.I.at the end of n years are given by nA=P(1+(R)/(100))^(n) and C.I.=A-P=P{(1+(R)/(100))^(n)-1} respectively.

Mohan purchased a colour TV. for Rs16,000 If the value of TV depreciates at the rate 5%0 per annum,then what will be the value of T.V,after 2 years?

Formula 3 Let P be the population of a city or a town at the beginning of a certain year. if the population decreases at the rate of R% per annum then population after n years = P (1- R/100)^n

If the rate of increase in population is r% per year, the population after n years is p, find the population that was n years before.

The amount(A) on Rs P for n years at the rate of r% compound interest per annum compounded at the interval of 3 months is

Formula 1 Let P be the population of a city or a town at the beginning of a certain year and the population grows at a constant rate of R% per annum then population after n years =P(1+(R)/(100))^(n)

Results on Depreciation : Value of the machine after n years ???? ???? ?? ????? n ????? ???